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Ksju [112]
3 years ago
13

If an electron loses energy what may happen to it

Physics
1 answer:
MAVERICK [17]3 years ago
7 0
<span>When an electron is hit by a photon of lights,it absorbs the quanta of energy the photon was carrying and moves to a higher energy state.Electrons therefore have to jump around within the atom as they either gain or lose energy.</span>
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Another name for the skin as a whole.
sergejj [24]
Organ, integument, dermis
5 0
3 years ago
Object A has a mass of 100 grams. Object B has a mass of 150 grams. They are both traveling at the same velocity. What can you c
Scilla [17]
-- Momentum is (mass) x (speed). 
Object B has 1.5 times as much momentum as Object A has.

-- Kinetic energy is (1/2) x (mass) x (speed) . 
Object B has 1.5 times as much kinetic energy as Object A has.

-- If they would both stop long enough to get on the scale,
Object B would weigh 1.5 times as much as Object A does.
8 0
3 years ago
A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp
Sergeu [11.5K]

Answer:

U=1.29\times 10^{-7}\ J

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm  ,K₁= 4.7

d₂=2.1 mm  , K₂=2.6

We know that capacitance given as

C_1=\dfrac{K_1\varepsilon _oA}{d_1}

C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}

C_1=1.26\times 10^{-10}\ F

C_2=\dfrac{K_2\varepsilon _oA}{d_2}

C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}

C_2=0.701\times 10^{-10}\ F

Net capacitance

C=\dfrac{C_1C_2}{C_1+C_2}

C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F

C=4.5\times 10^{-11}\ F

We know that stored energy given as

U=\dfrac{CV^2}{2}

V= 76 V

U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J

U=1.29\times 10^{-7}\ J

3 0
3 years ago
A 111 ‑turn circular coil of radius 2.11 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu
Ira Lisetskai [31]

Answer:

0.0061 J

Explanation:

Parameters given:

Number of turns, N = 111

Radius of turn, r = 2.11 cm = 0.0211 m

Resistance, R = 14.1 ohms

Time taken, t = 0.125 s

Initial magnetic field, Bin = 0.669 T

Final magnetic field, Bfin = 0 T

The energy dissipated in the resistor is given as:

E = P * t

Where P = Power dissipated in the resistor

Power, P, is given as:

P = V² / R

Hence, energy will be:

E = (V² * t) / R

To find the induced voltage (EMF), V:

EMF = [-(Bfin - Bin) * N * A] / t

A is Area of coil

EMF = [-(0 - 0.669) * 111 * pi * 0.0211²] / 0.125

EMF = 0.83 V

Hence, the energy dissipated will be:

E = (0.83² * 0.125) / 14.1

E = 0.0061 J

7 0
3 years ago
An arrow is shot at an angle of 35° and a velocity of 50 m/s. How long does it take to return to its original starting height?
Ostrovityanka [42]

Answer:

4.02 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 35°

Initial velocity (u) = 50 m/s

Acceleration due to gravity (g) = 10 m/s²

Time of flight (T) =?

The time of flight of the arrow can be obtained as follow:

T = 2uSineθ / g

T = 2 × 35 × Sine 35 / 10

T = 70 × 0.5736 / 10

T = 7 × 0.5736

T = 4.02 s

Therefore, the time taken for the arrow to return is 4.02 s

4 0
3 years ago
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